Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q^2 - q}{q^2 + 4q} \times \dfrac{-3q^2 - 15q - 12}{q^2 + 10q + 9} $
Solution: First factor out any common factors. $k = \dfrac{q(q - 1)}{q(q + 4)} \times \dfrac{-3(q^2 + 5q + 4)}{q^2 + 10q + 9} $ Then factor the quadratic expressions. $k = \dfrac {q(q - 1)} {q(q + 4)} \times \dfrac {-3(q + 1)(q + 4)} {(q + 1)(q + 9)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {q(q - 1) \times -3(q + 1)(q + 4) } {q(q + 4) \times (q + 1)(q + 9) } $ $k = \dfrac {-3q(q + 1)(q + 4)(q - 1)} {q(q + 1)(q + 9)(q + 4)} $ Notice that $(q + 1)$ and $(q + 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-3q\cancel{(q + 1)}(q + 4)(q - 1)} {q\cancel{(q + 1)}(q + 9)(q + 4)} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $k = \dfrac {-3q\cancel{(q + 1)}\cancel{(q + 4)}(q - 1)} {q\cancel{(q + 1)}(q + 9)\cancel{(q + 4)}} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $k = \dfrac {-3q(q - 1)} {q(q + 9)} $ $ k = \dfrac{-3(q - 1)}{q + 9}; q \neq -1; q \neq -4 $